A solution was made by combining 145 (±1) mL of 0.250 (±0.003) M NaCl with 275.0 (±0.5) mL of 6.00 (±0.01) wt% NaCl (assume this solution’s density is 1.0413 g/mL and the molar mass of NaCl is 58.443 g/mol, both with negligible error) in a 500.00 mL volumetric flask, and filling with water up to the line.
How many mmol of NaCl are in the final solution? Include absolute error in your number of millimoles and round your answer to an appropriate number of sig figs.
(HINT: First find the number of millimoles of NaCl, ignoring the error. Then go back and propagate error properly through each calculation step. )
First calcualte the number of moels in solution as follows:
145 (±1) mL of 0.250 (±0.003) M NaCl
Number of moles = volume * molarity
= 0.145 L * 0.250 M or moles / L
=0.03625 mol
275.0 (±0.5) mL of 6.00 (±0.01) wt% NaCl
density =1.0413 g/mL and the molar mass of NaCl = 58.443 g/mol,
Use the density to calculate the weight of 1 liter of
solution-
1000 mL x 1.0413 g/mL = 1041.3 grams
Use the percent to calculate the mass of NaCl in 1041.3 g (1.000
liters)-
1,041.3 x 6 % = 62.478 g NaCl
Now you know the weight of HCl per liter. Convert the weight of
NaCl to moles of HCl-
62.478 g / 58.443 g/mol = 1.069 moles of NaCl
Total number of moles = 0.03625 mol + 1.069 moles of NaCl
=1.1053 mol = 1105.3 mmol
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